Proof using properties of an isosceles or right-angle triangleGiven a △ A B C with sides A B = B C and ∠ B = 100 ∘ , prove that a 3 + b 3 = 3 a 2 b where a = A B = B C and b = A C.

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Proof using properties of an isosceles or right-angle triangleGiven a   △  A  B  C with sides   A  B  =  B  C and   ∠  B  =      100    ∘  , prove that       a    3    +      b    3    =  3      a    2    b where   a  =  A  B  =  B  C and   b  =  A  C.

Mohammad Orr · Accepted Answer

Step 1A straight forward application of cosine rule should tell you that   b  =  2  a  sin  ⁡  (  50  ).Step 2Consider                               a          3                +                  b          3                −        3                  a          2                b                            =                  a          3                (        1        +        8                  sin          3                ⁡        50        −        6        sin        ⁡        50        )                                          =                  a          3                (        1        +        8                              (            3            sin            ⁡            50            −            sin            ⁡            30            )                    4                −        6        sin        ⁡        50        )                                          =                  a          3                (        1        +        6        sin        ⁡        50        −        2        sin        ⁡        30        −        6        sin        ⁡        50        )                                          =        0

Kristen Garrison · Accepted Answer

Step 1      a    3    +      b    3    =  3      a    2    b using cosine      b    2    =      a    2    +      a    2    −  2  a  a  cos  ⁡  β  ,  β  &amp;lt;  100      b    2    =  2      a    2    (  1  −  cos  ⁡  β  )    l  e  t    (      b    a        )    2    =      t    2    =  2  (  1  −  cos  ⁡  β  )      a    3    +      b    3    =  3      a    2    b    ⟹    1  +      t    3    =  3  t    ⟹        t    3    −  3  t  +  1  =  0Step 2  t  (      t    2    −  3  )  =  −  1    ⟹        t    2    (      t    2    −  3      )    2    =  2  (  1  −  cos  ⁡  β  )  (  1  −  cos  ⁡  β  −  3      )    2      =  2  (  1  −  cos  ⁡  β  )  (  1  +  4  cos  ⁡  β  +  4      cos    2    ⁡  β  )    =  2  (  1  +  3  cos  ⁡  β  −  4      cos    3    ⁡  β  )  =  2  +  6  cos  ⁡  β  −  8      cos    3    ⁡  β  =  1Step 3Now we should show that   8      cos    3    ⁡  β  −  6  cos  ⁡  β  −  1  =  2  cos  ⁡  3  β  −  1  =  0  3  β  =  ±      π    3    +  2  π  k  ,  k  ∈  Z

Given a triangle ABC with sides AB=BC and angle B=100^circ, prove that a^3+b^3=3a^2b where a=AB=BC and b=AC

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